Question

In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the AP.

Answer 1

Answer:

Let a be the first term and d be the common difference of the given AP.

Sum of the first n terms is given by

Sn = n/2 {2a + (n - 1)d}

Putting n = 10, we get

S₁₀ = 10/2 {2a + (10 - 1)d}

210 = 5 (2a + 9d) 

2a + 9d = 210/5

2a + 9d = 42 ...............(1)

Sum of the last 15 terms is 2565

⇒ Sum of the first 50 terms - sum of the first 35 terms = 2565

S₅₀ - S₃₅ = 2565

⇒ 50/2 {2a + (50 - 1)d} - 35/2 {2a + (35 - 1)d} = 2565

25 (2a + 49d) - 35/2 (2a + 34d) = 2565

⇒ 5 (2a + 49d) - 7/2 (2a + 34d) = 513

⇒ 10a + 245d - 7a + 119d = 513

⇒ 3a + 126d = 513 

⇒ a + 42d = 171 ........(2)

Multiply the equation (2) with 2, we get

2a + 84d = 342 .........(3)

Subtracting (1) from (3)

  2a + 84d = 342

  2a + 9d   =  42

-      -         -

_______________

        75d = 300

_______________

 d= 4

Now, substituting the value of d in equation (1)

2a + 9d = 42

2a + 9*4 = 42

2a = 42 - 36

2a = 6

a = 3

So, the required AP is 3, 7, 11, 15, 19, 23, 27, 31, 35, 39 ........

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Answer 2

Step-by-step explanation:

Consider a and d as the first term and the common difference of an A.P. respectively.

n th term of an A.P., an = a + ( n – 1)d

Sum of n terms of an A.P., S n = n/ 2 [2a + (n – 1)d]

Given that the sum of the first 10 terms is 210.

⇒ 10 / 2 [2a + 9d ] = 210

⇒ 5[ 2a + 9 d ] = 210

⇒2a + 9d = 42 ----------- (1)

15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning

⇒ a36 = a + 35d

Sum of the last 15 terms = 15/2 [2a36 + ( 15 – 1)d ] = 2565

⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565

⇒ 15 [ a + 35d + 7d ] = 2565

⇒a + 42d = 171 ----------(2)

From (1) and (2), we have d = 4 and a = 3.

Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199.